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Mrrafil [7]
4 years ago
14

How does the structure of the alveoli relate to its function in the lungs

Physics
1 answer:
slavikrds [6]4 years ago
8 0
Structure of Alveoli provides more surface area for lungs, to exchange the gases & for process of respiration
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How to classify an inner planet vs. outer planet
zvonat [6]
Outer planets are farther away and made up of gases. Inner planets closer. It's pretty much self explanatory. Hope this helps.
7 0
4 years ago
What kind of injury would most likely come from an unfortunate slide into home in a baseball game?
quester [9]
Macule i do believe.
5 0
3 years ago
If a dog has a potential energy of 1500 J and a mass of 24 kg, how high off the ground is he?
Llana [10]

Answer:

h = 6.371 m

Explanation:

Data:

Potential Energy = P.E = 1500 J

Mass = m = 24 Kg

Gravity = g = 9.81 m/sec²

Height = h = ?

Solution:

As the formula for the Potential Energy is;

P.E = mgh

Replacing the values in the above equation

1500 = 24 × 9.81 × h

h = (1500)÷(24 × 9.81)

h = 6.371 m

8 0
3 years ago
An organic compound is a compound that usually contains carbon and hydrogen, and that is found in living organisms. True or Fals
Assoli18 [71]

This statement is true. Organic compounds hold carbon and hydrogen along with oxygen and nitrogen.

7 0
4 years ago
Imagine an alternative universe where the characteristic decay time of neutrons is 3 min instead of 15 min. All other properties
FrozenT [24]

Answer:

(a) [Y_{p} ]_{max} = \frac{2f}{1+f}

(b) f_{new} = 0.013; [Y_{p} ]_{max} = 0.026

Explanation:

Since the neutron-to-proton ratio at the time of nucleosynthesis is given:

f = \frac{n_{n} }{n_{p} }

Therefore:

n_{n} = f*n_{p}

Then, to determine the maximum ⁴He fraction if all the available n_{n} neutrons bind to all the protons. Since, there are 2 protons and 2 neutrons in a ⁴He nucleus, it shows that there would be n_{n}/2 nuclei of ⁴He.

In addition, a ⁴He nucleus has a mass of 4m_{p}, where m_{p} is the mass of one proton. Thus, n_{n}/2 nuclei of such nuclei will have a mass of n_{n}/2*4m_{p}.

Assuming that m_{p}=m_{n}, there would be a total of (n_{n}+n_{p}) protons and neutrons with a total mass of (n_{n}+n_{p})*m_{p}.

Thus:[Y_{p} ]_{max} = \frac{2f}{1+f}

(b) Given:

t_{nuc} = 200 s;   τ_{n} = 3*60s = 180 s

f_{new} = \frac{n_{nf} }{n_{pf} } = \frac{exp (-200/180)}{5 +[1- exp(-200/180)]} =\frac{0.077}{5.923} = 0.013

[Y_{p} ]_{max} = \frac{2f}{1+f} = (2*0.013)/(1+0.013) = 0.026

6 0
3 years ago
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